题目：

sql建表语句：

Create table If Not Exists Customers (customer_id int, customer_name varchar(20), email varchar(30));
Create table If Not Exists Contacts (user_id int, contact_name varchar(20), contact_email varchar(30));
Create table If Not Exists Invoices (invoice_id int, price int, user_id int);
Truncate table Customers;
insert into Customers (customer_id, customer_name, email) values ('1', 'Alice', 'alice@leetcode.com');
insert into Customers (customer_id, customer_name, email) values ('2', 'Bob', 'bob@leetcode.com');
insert into Customers (customer_id, customer_name, email) values ('13', 'John', 'john@leetcode.com');
insert into Customers (customer_id, customer_name, email) values ('6', 'Alex', 'alex@leetcode.com');
Truncate table Contacts;
insert into Contacts (user_id, contact_name, contact_email) values ('1', 'Bob', 'bob@leetcode.com');
insert into Contacts (user_id, contact_name, contact_email) values ('1', 'John', 'john@leetcode.com');
insert into Contacts (user_id, contact_name, contact_email) values ('1', 'Jal', 'jal@leetcode.com');
insert into Contacts (user_id, contact_name, contact_email) values ('2', 'Omar', 'omar@leetcode.com');
insert into Contacts (user_id, contact_name, contact_email) values ('2', 'Meir', 'meir@leetcode.com');
insert into Contacts (user_id, contact_name, contact_email) values ('6', 'Alice', 'alice@leetcode.com');
Truncate table Invoices;
insert into Invoices (invoice_id, price, user_id) values ('77', '100', '1');
insert into Invoices (invoice_id, price, user_id) values ('88', '200', '1');
insert into Invoices (invoice_id, price, user_id) values ('99', '300', '2');
insert into Invoices (invoice_id, price, user_id) values ('66', '400', '2');
insert into Invoices (invoice_id, price, user_id) values ('55', '500', '13');
insert into Invoices (invoice_id, price, user_id) values ('44', '60', '6');

分析，这道题最难的就是理解题目，该客户的联系人，和可信联系人，这两个非常难理解，我认为该客户的联系人就是Contacts 表中的所有联系人都是，可信联系人是指Contacts 表中邮箱在客户表中的人，这里我们首先先把Contacts 中的可信联系人标记出来，如果是可信联系人就标记为1 否则标记为0，然后分组求出每个人的联系人，还有每个人的可信联系人，全部联系人，这里是用count算出，可信联系人，这里是用sum标记列算出，然后连接三个表就得出来结果了，图表分析：

sql实现：

with t1 as (select *, if(contact_email in (select email from Customers), 1, 0) ll from Contacts),       -- 首先先给可信人员打上标签，可信人返回1 不可信人返回0
     t2 as (select invoice_id, price, user_id, customer_name
            from Invoices i1,
                 Customers c1
            where i1.user_id = c1.customer_id),   -- 然后连接发票表和用户表
     t3 as (select user_id, count(1) contacts_cnt, sum(ll) trusted_contacts_cnt from t1 group by user_id)  -- 然后算出所有联系人和可信联系人的人数
select invoice_id,
       customer_name,
       price,
       ifnull(contacts_cnt, 0)         contacts_cnt,
       ifnull(trusted_contacts_cnt, 0) trusted_contacts_cnt
from t2
         left join t3 on t2.user_id = t3.user_id order by invoice_id  -- 连接发票表和可信人数表然后人数如果为空，就返回零，否则返回人数，按照发票id排序

pandas例子：

data = [[1, 'Alice', 'alice@leetcode.com'], [2, 'Bob', 'bob@leetcode.com'], [13, 'John', 'john@leetcode.com'], [6, 'Alex', 'alex@leetcode.com']]
customers = pd.DataFrame(data, columns=['customer_id', 'customer_name', 'email']).astype({'customer_id':'Int64', 'customer_name':'object', 'email':'object'})
data = [[1, 'Bob', 'bob@leetcode.com'], [1, 'John', 'john@leetcode.com'], [1, 'Jal', 'jal@leetcode.com'], [2, 'Omar', 'omar@leetcode.com'], [2, 'Meir', 'meir@leetcode.com'], [6, 'Alice', 'alice@leetcode.com']]
contacts = pd.DataFrame(data, columns=['user_id', 'contact_name', 'contact_email']).astype({'user_id':'Int64', 'contact_name':'object', 'contact_email':'object'})
data = [[77, 100, 1], [88, 200, 1], [99, 300, 2], [66, 400, 2], [55, 500, 13], [44, 60, 6]]
invoices = pd.DataFrame(data, columns=['invoice_id', 'price', 'user_id']).astype({'invoice_id':'Int64', 'price':'Int64', 'user_id':'Int64'})

pandas分析：大致与sql分析一样

pandas实现：

import pandas as pd

def count_trusted_contacts(customers: pd.DataFrame, contacts: pd.DataFrame, invoices: pd.DataFrame) -> pd.DataFrame:
    contacts['cn']=contacts['contact_email'].apply(lambda x:1 if x in customers['email'].values else 0) -- 给联系表中的人打标签，如果是可信人返回1 否则返回零

    d=contacts.groupby('user_id').agg({'contact_name':'count','cn':'sum'}).reset_index()  -- 算出每个人的全部联系人和可信人数

    b=invoices.merge(customers,left_on='user_id',right_on='customer_id',how='left')  -- 连接发票表和客户表

    c=b.merge(d,left_on='user_id',right_on='user_id',how='left') -- 连接发票姓名表和联系人数和可信人数表
    c=c[['invoice_id','customer_name','price','contact_name','cn']].fillna(0).sort_values('invoice_id')  -- 把空值设置成0，然后排序

    c.columns=['invoice_id','customer_name','price','contacts_cnt','trusted_contacts_cnt'] -- 修改列名
    return c